对于二叉树的层序遍历，BFS方法是更为常用的思路。但我觉得用DFS递归的方法做也很好，下面贴出代码：

class Solution:
    def levelOrder(self, root: TreeNode) -> List[List[int]]:
        self.levelList = []
        if not root:
            return []
        self.dfs(root, 0)
        return self.levelList

    def dfs(self, root, level):
        if not root:
            return
        if len(self.levelList) < level + 1:
            self.levelList.append([])
        self.levelList[level].append(root.val)
        
        self.dfs(root.left, level+1)
        self.dfs(root.right, level+1)